Question

Se da determinantul ca in următoarea poza:
Sa se arate ca detA=(a-b)(b-c)(c-a)​

Answer 1

$\left|\begin{array}{ccc}1&a&a^2\\1&b&b^2\\1&c&c^2\end{array}\right|\overset{ \begin{array}{lcl}L_2 \to L_2-L_1\\L_3 \to L_3-L_1\end{array}}{=\joinrel=} \left|\begin{array}{ccc}1&a&a^2\\0&b-a&(b-a)(b+a)\\0&c-a&(c-a)(c+a)\end{array}\right| = \\ \\ \\= (b-a)(c-a)\left|\begin{array}{ccc}1&a&a^2\\0&1&b+a\\0&1&c+a\end{array}\right| = \\ \\\\ = (b-a)(c-a)\Big[c+a-(b+a)\Big] = \\ \\ = (b-a)(c-a)(c-b) = \\ \\ = (a-b)(b-c)(c-a)$